Why Moving Endpoints Matter
An integral can describe accumulated change up to a chosen endpoint. If that endpoint depends on \(x\), then changing \(x\) changes the accumulation limit as well. The derivative must therefore account for how quickly the endpoint moves.
For a polynomial integrand, its value at the moving endpoint gives the rate of accumulation there. The chain rule then accounts for the endpoint's rate of change. If the lower endpoint moves instead, the sign is negative because increasing the lower limit removes area from the integral. When both endpoints move, both effects contribute.
This tutorial develops these rules for polynomial integrands and works through upper limits, lower limits, and two moving endpoints. The computations use the Fundamental Theorem of Calculus and the chain rule.
The Rule for a Moving Upper Endpoint
Let \(p\) be a polynomial, let \(c\) be a constant, and let \(g\) be differentiable. Define an accumulation function by integrating \(p\) from \(c\) to \(g(x)\). Since a polynomial is continuous everywhere, the Fundamental Theorem of Calculus applies. The chain rule then gives the derivative.
The result has two parts. First, evaluate the integrand at the endpoint, \(p(g(x))\). Then multiply by the endpoint's derivative, \(g'(x)\). The endpoint value alone is the derivative with respect to the endpoint, not necessarily with respect to \(x\).
Worked Example: A Nonlinear Upper Endpoint
Differentiate \[ F(x)=\int_1^{x^2+1}(3t^2-2t+4)\,dt. \]
Here \(p(t)=3t^2-2t+4\) and \(g(x)=x^2+1\), so \(g'(x)=2x\). Apply the moving-upper-endpoint rule:
Check: An antiderivative of the integrand is \(P(t)=t^3-t^2+4t\). Thus \(F(x)=P(x^2+1)-P(1)\). Differentiating this expression gives \[ P'(x^2+1)(2x) = \bigl(3(x^2+1)^2-2(x^2+1)+4\bigr)(2x) = 6x^5+8x^3+10x, \] which agrees with the result.
Why the Chain-Rule Factor Appears
The rule follows directly by expressing the variable-limit integral as a composition. Choose a fixed reference point, say \(0\), and define \(A(u)=\int_0^u p(t)\,dt\). Additivity of definite integrals gives \[ \int_c^{g(x)}p(t)\,dt=A(g(x))-A(c). \] By the Fundamental Theorem of Calculus, \(A'(u)=p(u)\), because \(p\) is continuous. Since \(A(c)\) is constant, differentiating the right-hand side uses the chain rule:
This also explains why the derivative has the displayed form: changing \(x\) changes the accumulated value by changing \(g(x)\), and the chain rule multiplies the rate of accumulation at that endpoint by the rate at which the endpoint moves.
A Moving Lower Endpoint
A lower endpoint changes the integral in the opposite direction. Increasing the lower limit excludes a portion of the accumulated interval. The negative sign in the rule records this effect. It also follows from reversing the bounds: \(\int_{g(x)}^c p(t)\,dt=-\int_c^{g(x)}p(t)\,dt\).
Worked Example: A Moving Lower Endpoint
Differentiate \[ G(x)=\int_{x-1}^{5}(2t^3-3t+1)\,dt. \]
The lower endpoint is \(g(x)=x-1\), with \(g'(x)=1\). The integrand is \(p(t)=2t^3-3t+1\). Therefore \[ G'(x)=-p(x-1)(1). \] Evaluate the polynomial at the endpoint step by step:
Hence \[ G'(x)=-(2x^3-6x^2+3x+2) =-2x^3+6x^2-3x-2. \]
Check: An antiderivative of \(2t^3-3t+1\) is \(P(t)=\frac{t^4}{2}-\frac{3t^2}{2}+t\). Therefore \(G(x)=P(5)-P(x-1)\), and differentiating gives \[ G'(x)=-P'(x-1)(1) =-(2(x-1)^3-3(x-1)+1) =-2x^3+6x^2-3x-2. \] This matches the result from the lower-endpoint rule.
Two Moving Endpoints
When both limits depend on \(x\), the integral can be written as the difference of two accumulation functions. The upper endpoint contributes positively, and the lower endpoint contributes negatively. Both contributions include their own chain-rule factors.
To see why, fix a constant \(c\) and write \[ \int_{a(x)}^{b(x)}p(t)\,dt = \int_c^{b(x)}p(t)\,dt-\int_c^{a(x)}p(t)\,dt. \] Apply the upper-endpoint rule to each term. The first derivative is \(p(b(x))b'(x)\); the second is \(p(a(x))a'(x)\), and it is subtracted. This proves the formula under the stated continuity and differentiability conditions.
Worked Example: One Linear and One Nonlinear Endpoint
Differentiate \[ H(x)=\int_x^{x^2}(t^2+2t)\,dt. \]
Set \(p(t)=t^2+2t\), \(a(x)=x\), and \(b(x)=x^2\). Then \(a'(x)=1\) and \(b'(x)=2x\). Substitute each endpoint into the rule:
Check: An antiderivative is \(P(t)=\frac{t^3}{3}+t^2\). Evaluating the integral gives \[ H(x)=P(x^2)-P(x) =\frac{x^6}{3}+x^4-\frac{x^3}{3}-x^2. \] Differentiating term by term, \[ H'(x)=2x^5+4x^3-x^2-2x, \] as before.
A Sign Pitfall and How to Avoid It
A common error is to treat a moving lower endpoint as though it were an upper endpoint. For example, for the function \(G\) above, an incorrect calculation might be \[ G'(x)=p(x-1)=2x^3-6x^2+3x+2. \] This misses the fact that \(x-1\) is the lower limit. The correct calculation is \[ G'(x)=-p(x-1)=-2x^3+6x^2-3x-2. \] The antiderivative check confirms the negative sign: \(G(x)=P(5)-P(x-1)\), so the variable term is subtracted before differentiation.
For two moving endpoints, write the upper contribution first and the lower contribution second: \[ p(b(x))b'(x)-p(a(x))a'(x). \] Keeping this order visible helps prevent a sign error. It also makes clear that each endpoint has its own derivative.
A Reliable Calculation Process
For a polynomial integrand, the following sequence organizes the work and makes the result easier to check.
Write the polynomial as \(p(t)\), and label the lower and upper limits separately.
Find the derivative of each limit with respect to \(x\); a constant endpoint has derivative zero.
Use the upper endpoint in the positive term and the lower endpoint in the negative term.
Expand carefully, then evaluate the integral using an antiderivative and differentiate that expression to compare.
Why These Rules Are Useful
Variable-limit derivatives connect accumulated quantities to instantaneous rates. If a polynomial describes a rate, integrating it over an interval gives accumulated change on that interval. Differentiating with respect to a changing endpoint tells how the accumulation changes as the interval shifts or grows.
The endpoint rules also prevent unnecessary work. One can differentiate the integral directly from the integrand and the endpoint functions, without first expanding a potentially lengthy antiderivative expression. The antiderivative remains useful as an independent check, especially when endpoint substitutions produce several terms.
Keep three details in view: evaluate the polynomial at the endpoint, multiply by that endpoint's derivative, and use a negative sign for the lower endpoint. For two moving limits, apply these steps to each endpoint and combine the contributions as upper minus lower.
Check Your Understanding
For each problem, identify the endpoint derivatives before simplifying. Use an antiderivative to check at least one result.
- Differentiate \(\int_0^{3x+1}(t^2+1)\,dt\).
- Differentiate \(\int_{x^2}^{4}(t^3-2t)\,dt\).
- Differentiate \(\int_{2x}^{x+1}(t^2+3)\,dt\).
- For a moving lower limit \(a(x)\), explain why its contribution has a negative sign.
- In the two-endpoint rule, which endpoint is substituted into the positive term, and what factor accompanies it?
